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\begin{document}
	
\pagestyle{fancy}
\fancyhead{}
\lhead{Chenyue (22035026)}
\chead{EXE8}
\rhead{2021/5/12}


\section*{exe8.14:}

\begin{figure}[htb] 
	
	
	\center{\includegraphics[width=13cm]  {NA_exe8_14_1.png}} 
	
	
	\caption{\label{1} error and the order of accuracy} 
	
	
\end{figure}

\begin{figure}[htb] 
	
	
	\center{\includegraphics[width=8cm]  {NA_exe8_14_2.png}} 
	
	
	\caption{\label{1} geometric interpertation} 
	
	
\end{figure}


\section*{exe8.15:}

\[
\begin{aligned}
	E(D_{+}u(x))&=D_{+}u(x)-u^{'}(x)\\
	            &=\dfrac{u(x+h)-u(x)}{h}-u^{'}(x)\\
	            &=\dfrac{u^{'}(x)h+0.5u^{(2)}(x)h^2+o(h^2)}{h}-u^{'}(x)\\
	            &=0.5u^{(2)}(x)h+o(h)\\
	            &=O(h)       
\end{aligned}            
\]
可见$D_{+}u(x)$是一阶精度的。

\[
	\begin{aligned}
		E(D_{-}u(x))&=D_{-}u(x)-u^{'}(x)\\
		            &=\dfrac{u(x)-u(x-h)}{h}-u^{'}(x)\\
		            &=\dfrac{u^{'}(x)h-0.5u^{(2)}(x)h^2+o(h^2)}{h}-u^{'}(x)\\
		            &=-0.5u^{(2)}(x)h+o(h)\\
		            &=O(h) 
	\end{aligned}
\]
可见$D_{-}u(x)$是一阶精度的。

\[
	\begin{aligned}
		E(D_{0}u(x))&=D_{0}u(x)-u^{'}(x)\\
		            &=\dfrac{u(x+h)-u(x-h)}{2h}-u^{'}(x)\\
		            &=\dfrac{u(x)+u^{'}(x)h+0.5u^{(2)}(x)h^2+\dfrac{u^{(3)}(x)h^3}{6}+O(h^4)-u(x)+u^{'}(x)h-0.5u^{(2)}(x)h^2+\dfrac{u^{(3)}(x)h^3}{6}+O(h^4)}{2h}-u^{'}(x)\\
		            &=\dfrac{u^{(3)}(x)h^2}{6}+O(h^3)\\
		            &=O(h^2)
	\end{aligned}
\]
可见$D_{0}u(x)$是二阶精度的。

\section*{exe8.16:}
差分表如右所示：
\begin{tabular}{c|ccc}
$\bar{x}-2h$&$u(\bar{x}-2h)$& & \\
$\bar{x}-h$&$u(\bar{x}-h)$&$\dfrac{u(\bar{x}-h)-u(\bar{x}-2h)}{h}$ & \\
$\bar{x}$&$u(\bar{x})$&$\dfrac{u(\bar{x})-u(\bar{x}-h)}{h}$ &$\dfrac{u(\bar{x})-2u(\bar{x}-h)+u(\bar{x}-2h)}{2h^2}$ \\	
\end{tabular}
\[
	\begin{aligned}
	\therefore u(x)&=u(\bar{x}-2h)+\dfrac{u(\bar{x}-h)-u(\bar{x}-2h)}{h}(x-\bar{x}+2h)+\dfrac{u(\bar{x})-2u(\bar{x}-h)+u(\bar{x}-2h)}{2h^2}(x-\bar{x}+2h)(x-\bar{x}+h)\\
	\therefore
	u^{'}(x)&=\dfrac{u(\bar{x}-h)-u(\bar{x}-2h)}{h}+\dfrac{u(\bar{x})-2u(\bar{x}-h)+u(\bar{x}-2h)}{2h^2}(2x-2\bar{x}+3h)\\
	\therefore
	u^{'}(\bar{x})&=\dfrac{3u(\bar{x})-4u(\bar{x}-h)+u(\bar{x}-2h)}{2h}
    \end{aligned}
\]
结果和式子（8.19）一样。

\section*{exe8.20:}
\[
	\begin{aligned}
		&\vert\vert \mathbf{g} \vert\vert _{\infty}=\underset{1 \leq i \leq N}{max}|g_{i}|=O(h)\\
		&\vert\vert \mathbf{g} \vert\vert _{1}=h\sum_{n=1}^{N}|g_{i}|=h(2O(h)+(N-2)O(h^2))=O(h^2)\\
		&\vert\vert \mathbf{g} \vert\vert _{2}=(h\sum_{n=1}^{N}|g_{i}|^{2})^{\dfrac{1}{2}}=(h(2O(h^2)+(N-2)O(h^{4}))^{\dfrac{1}{2}}=(O(h^3))^{\dfrac{1}{2}}=O(h^{\dfrac{3}{2}})
	\end{aligned}
\]
得证。

\section*{exe8.44:}
设$B_{E}$的第一列为：
\[
\begin{bmatrix}
	b_{11}\\
	b_{21}\\
	\vdots\\
	b_{(m+1)1}\\
	b_{(m+2)1}	
\end{bmatrix}
\]
由于$A_{E}B_{E}=I$，所以得到方程组：
\[ \left\{
\begin{aligned}
	b_{21}-b_{11} & =h\\
	b_{11}-2b_{21}+b_{31}& = 0\\
	&.\\
	&.\\
	&.\\
	b_{(m)1}-2b_{(m+1)1}+b_{(m+2)1}&=0\\
	b_{(m+2)1}&=0
\end{aligned}
\right.
\]
所以求得$B_{E}$的第一列为：
\[
\begin{bmatrix}
	(m+1)h\\
	mh\\
	\vdots\\
	h\\
	0	
\end{bmatrix}
\]
从而可知$B_{E}$的第一列是$O(1)$的。

\end{document}

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